A parallel–plate capacitor is initially charged by connecting it to a battery. The battery is then disconnected. If the distance between the plates is increased, what happens to the charge on the capacitor and the voltage across it? a. The charge remains fixed and the voltage decreases.
b. The charge decreases and the voltage remains fixed.
c. The charge remains fixed and the voltage increases.
d. The charge decreases and the voltage increases.

t the battery of potential difference  V be used to charge the capacitor of capacitance  C.

∴ the charge stored in the capacitor      q=CV

Now the battery is disconnected, so the the charge  of the capacitor becomes constant

i.e    q=constant     OR     CV=constant                (1)

Capacitance of parallel plate capacitor        C=  

d

Aϵ  

o

So if the distance between the plates is increased, then the capacitance will decrease which is compensated by the increase  in voltage across the capacitor according to equation (1).

Also the energy stored in the capacitor          E=  

2C

q  

2

⟹E∝  

C

1

                (∵q=constant)

Thus energy will increase due to the decrease in capacitance.

Explanation:

less than one you can't create energy